Steel Compression Member φPn

ϕ P_{n} = ϕ F_{cr} A_{g}, λ = \frac{K L}{r π} \frac{F _{y}}{E}

AISC: inelastic vs elastic buckling split at λ=1.5 (handbook form). Use the governing (largest) KL/r. φ=0.90 (LRFD). Aligns with Euler when long and slender.

$K L / r$: Governing slenderness
$F cr$: Critical stress
$A g$: Gross area

Step-by-step solved example

Long column, Fcr from Euler 50 ksi, Ag=10 in², φ=0.9. φPn?

Steel column: KL/r and Fcr Ag. K from the end sketch.

1. Pn
Fcr Ag=500 kip.
2. φPn
0.9×500=450 kip.

Answer: φPn = 450 kip

10 practice questions

0/10 correct

1.K for a pinned–pinned column is

2.Fixed–free (flagpole) K is

3.Least r = √(Imin/Ag)

4.E of steel on FE is

5.Inelastic (stocky) Fcr is less than Fy but

6.Bracing that cuts unbraced L in half

7.ASD uses Ω=1.67 so Pn/Ω vs φPn=0.90 Pn is

8.KL/r=200, E=29000, Fy=50: Euler Fcr= π²E/(KL/r)² nearest

9.Recommended max KL/r for steel columns is often

10.Ag for a W shape is